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# Determining the bits in the offset of a logical and physical address.

Posted on 2005-05-13
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I need help in determining the bits in the offset of a logical and physical address and determining bytes in a page.
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Question by:imy2000
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Expert Comment

ID: 14001509
If your offset is k bits and your logical address is n bits, then you can have maximum page size of 2^k addressable locations (bytes).

Putting it in another way, given a page size of 2^k you need k bits in the logical address to denote the offset within a page.

If you have some specific question or need some further clarification, feel free to post back
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Author Comment

ID: 14002433
A computer has 12-bit addresses and a main memory that is divided into 16 frames.  How many bytes are in each page?How many bits are in the offset of a logical and physical address?
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Expert Comment

ID: 14002939
Is this an assignment?
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Author Comment

ID: 14003676
does it matter?
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Expert Comment

ID: 14004535
Yes
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Accepted Solution

sunnycoder earned 200 total points
ID: 14004540
What is the maximum amount of memory you can address with 12 bits?

>and a main memory that is divided into 16 frames. How many bytes are in each page?
Divide this maximum memory into 16 equal frames
How much memory would make up a frame

>How many bits are in the offset of a logical and physical address?
given a page size of 2^k you need k bits in the logical address to denote the offset within a page.
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